What is the time complexity of Minimum Cost of Buying Candies With Discount?

The optimal solution has a time complexity of O(n log n).

What data structures are used to solve Minimum Cost of Buying Candies With Discount?

This problem uses Array, Greedy, Sorting. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Minimum Cost of Buying Candies With Discount?

Insight 1: Sorting the candies in descending order allows us to efficiently pick the most expensive candies first, ensuring we get the maximum discount. Insight 2: A greedy approach works optimally here. By always selecting the most expensive candies to pay for and getting the cheapest as a freebie, we minimize the overall cost.

Which companies ask LeetCode 2144. Minimum Cost of Buying Candies With Discount in interviews?

This problem is frequently asked at Garmin, Nokia.

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Minimum Cost of Buying Candies With Discount - LeetCode 2144 Solution

Minimum Cost of Buying Candies With Discount - Complete Solution Guide

Minimum Cost of Buying Candies With Discount is LeetCode problem 2144, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

A shop is selling candies at a discount. For every two candies sold, the shop gives a third candy for free . The customer can choose any candy to take away for free as long as the cost of the chosen candy is less than or equal to the minimum cost of the two candies bought. For example, if there are 4 candies with costs 1 , 2 , 3 , and 4 , and the customer buys candies with costs 2 and 3 , they can take the candy with cost 1 for free, but not the candy with cost 4 . Given a 0-indexed integer arra

Detailed Explanation

The problem asks you to determine the minimum cost to buy all candies given a list of candy prices. The twist is that for every two candies purchased, a third candy (the cheapest among the two purchased and remaining candies) is given for free. The input is a 0-indexed integer array `cost` where `cost[i]` represents the price of the i-th candy. The output is a single integer representing the minimum total cost to acquire all candies.

Solution Approach

The solution utilizes a greedy strategy combined with sorting. The candies are first sorted in descending order of price. Then, the algorithm iterates through the sorted array, picking the first two candies in each group of three as paid items and considering the third as free. This process minimizes the total cost by always choosing the most expensive candies to purchase.

Step-by-Step Algorithm

Step 1: Sort the `cost` array in descending order.
Step 2: Iterate through the sorted array with a step of 3 (i.e., 0, 3, 6...).
Step 3: For each iteration, add the cost of the current candy (`cost[i]`) and the next candy (`cost[i+1]`, if it exists) to the `totalCost`.
Step 4: After the loop, return `totalCost`.

Key Insights

Insight 1: Sorting the candies in descending order allows us to efficiently pick the most expensive candies first, ensuring we get the maximum discount.
Insight 2: A greedy approach works optimally here. By always selecting the most expensive candies to pay for and getting the cheapest as a freebie, we minimize the overall cost.
Insight 3: Handling the edge case of having fewer than 3 candies requires handling the remainder after grouping the candies in sets of 3.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Greedy, Sorting.

Companies

Asked at: Garmin, Nokia.

Minimum Cost of Buying Candies With Discount | LeetCode 2144

Easy

Array Greedy Sorting

Time: O(n log n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int minimumCost(int[] cost) {
        Arrays.sort(cost);
        int n = cost.length;
        int totalCost = 0;
        for (int i = n - 1; i >= 0; i -= 3) {
            totalCost += cost[i];
            if (i - 1 >= 0) {
                totalCost += cost[i - 1];
            }
        }
        return totalCost;
    }
}

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Problem Info

Description

Problem 2144: Minimum Cost of Buying Candies With Discount - Detailed explanation and solution approaches.

Topics

Array Greedy Sorting

Companies

GarminNokia

Stats

Difficulty:Easy

Languages:4

Companies:2

Problem #:2144

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

4 steps

1
Step 1: Sort the `cost` array in descending order.
2
Step 2: Iterate through the sorted array with a step of 3 (i.e., 0, 3, 6...).
3
Step 3: For each iteration, add the cost of the current candy (`cost[i]`) and the next candy (`cost[i+1]`, if it exists) to the `totalCost`.
4
Step 4: After the loop, return `totalCost`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The dominant operation is the sorting of the `cost` array, which takes O(n log n) time using a comparison-based sorting algorithm like merge sort or quicksort. The subsequent iteration through the array takes O(n) time, but this is dominated by the sorting. The space complexity is O(1) because the algorithm uses a constant amount of extra space regardless of the input size. In-place sorting can further optimize space usage.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Iterative without Sorting

Instead of sorting, iteratively find the maximum, second maximum, and minimum elements. This involves multiple scans of the array.

Time: O(n^2)Space: O(1)

Trade-offs: This approach avoids the sorting step but increases the time complexity significantly. It's less efficient than the sorting-based approach, especially for larger input arrays.

✨

Key Insights

1Insight 1: Sorting the candies in descending order allows us to efficiently pick the most expensive candies first, ensuring we get the maximum discount.
2Insight 2: A greedy approach works optimally here. By always selecting the most expensive candies to pay for and getting the cheapest as a freebie, we minimize the overall cost.
3Insight 3: Handling the edge case of having fewer than 3 candies requires handling the remainder after grouping the candies in sets of 3.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: Empty input array. The provided solutions handle this implicitly by either having a check (C code) or by natural behavior of the loop (Python, Java, C++). The total cost would be 0.
Edge case 2: Array with fewer than 3 elements. The algorithm correctly handles this by simply summing the costs of all candies in the array.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to sort the array before iterating. This will lead to incorrect results because the greedy approach relies on the sorted order to pick the optimal candies for free.

✗Common mistake 2: Incorrectly handling the edge case of an array with fewer than 3 elements, leading to incorrect calculation of the total cost.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Optimizing bulk purchases with discounts. Many stores offer discounts for buying items in bulk. This algorithm can be adapted to determine the most cost-effective way to make such purchases.

🏦Real-world scenario 2: Resource allocation. Imagine allocating resources where some are free or come at a reduced cost if others are utilized. This algorithm can help find the optimal allocation to minimize total cost.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Don't rush. Explain your thought process clearly. Easy problems are often about communication.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: November 12, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I verify if a greedy approach is correct?

Prove the greedy choice property: the locally optimal choice leads to globally optimal solution. Use exchange argument: show that swapping any element with the greedy choice doesn't improve the solution. If you can't prove it, test extensively or consider DP.

How should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Cost of Buying Candies With Discount"?

For this easy-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether greedy techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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